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Let $u$ be a solution of $\partial_t u - \Delta u =f$ with initial data $u(0,x) = 0$ on $\mathbb R^N$. How do one prove the following inequality?

$$ \int_0^T \int_{\mathbb R^N} \phi(f)(- \Delta) u(s,x)dxds \ge 0, $$ where $\phi$ is a smooth non-decreasing function.

How can we estimate $$\int_0^T \int_{\mathbb R^N} \phi(f) (-\Delta )u(s,x)dxds$$ from below more precisely?

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  • $\begingroup$ Why this should be true? $\endgroup$
    – Giorgio Metafune
    Commented Mar 24, 2020 at 9:10
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    $\begingroup$ When $\phi(t)=t$ the converse inequality is true. $\endgroup$
    – Giorgio Metafune
    Commented Mar 24, 2020 at 9:31
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    $\begingroup$ @GiorgioMetafune I don't know why it should be true, but I wish it to be the case. There was a typo and I meant $-\Delta$ in the integral. $\endgroup$
    – user140746
    Commented Mar 24, 2020 at 10:10
  • $\begingroup$ Thank you, I see now. $\endgroup$
    – Giorgio Metafune
    Commented Mar 24, 2020 at 10:19
  • $\begingroup$ Do you know what happens for the elliptic counterpart, that is for $u-\Delta u=f$? Or for specific $\phi?$. Thank you $\endgroup$
    – Giorgio Metafune
    Commented Mar 24, 2020 at 15:32

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